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Section 3.5 Implicit Differentiation

In the previous section, we computed the derivative of \(y=\ln x\) using the fact that it is the inverse of \(e^x\) by writing an equation for the inverse in terms of the original function [\(e^y=x\)] and using the Chain Rule to get a formula for \(y'\) in terms of the derivative of the original function. This is one case of the strategy of implicit differentiation, where the function to be differentiated is given implicitly as the solution of an equation, rather than by an explicit formula. Here we will see other uses for this strategy, like computing the slope at a point on a curve when the curve is given by an equation, not as the graph of a known function.
Do this two ways:
  1. By finding an explicit equation \(y=f(x)\) for the curve.
  2. by differentiating the equation
    \begin{equation*} x^2 + [f(x)]^2 = 25 \end{equation*}
    without using a formula for \(f(x)\text{.}\)
Find the tangent line to the point \(P(1,1)\) on the curve \(2x^7+y^7=3xy\text{.}\)
Procedure for Implicit Differentiation, with example \(x^2+xy^2+e^y = 8\).

To find a formula for the derivative of a function \(y=f(x)\) given implicitly by an equation involving \(x\) and \(y\text{:}\)

  1. Differentiate each side of the equation.

    Note that every time \(y\) appears, you must use the Chain Rule.

    \begin{equation*} 2x + \left( y^2 + x2y \cdot \frac{dy}{dx} \right) + e^y \cdot \frac{dy}{dx} = \frac{d}{dx}8 = 0. \end{equation*}

    (The parentheses are around the derivative of \(xy^2\text{,}\) which also requires the product rule.)

  2. Add and subtract to get terms with factor \(\displaystyle\frac{dy}{dx}\) at left, all others at right.
    \begin{equation*} 2xy \cdot \frac{dy}{dx}+ e^y \cdot \frac{dy}{dx} = -2x-y^2. \end{equation*}
  3. Collect the common factor \(\displaystyle\frac{dy}{dx}\) present in every term at left.
    \begin{equation*} \frac{dy}{dx}(2xy + e^y) = -(2x+y^2). \end{equation*}
  4. Divide out to get a formula for \(\displaystyle\frac{dy}{dx}\text{,}\) which is the desired answer.
    \begin{equation*} \frac{dy}{dx} = -\frac{2x+y^2}{2xy + e^y}. \end{equation*}

Note that to use this formula, you need both coordinates of a point of the curve given by the original equation, \(P(x_0,y_0)\text{,}\) not just an \(x\) value.

Hint
Note that the “hidden” Chain Rule comes up in two places, \(\displaystyle\frac{d}{dx}(y^2)\) and \(\displaystyle\frac{d}{dx}(x+y)\text{.}\)
Solution
  1. The chain rule at left and product rule at right give:
    \begin{equation*} \cos(x+y)\frac{d}{dx}(x+y) = \frac{d}{dx}(y^2) \cos x + y^2 \frac{d}{dx}(\cos x), \end{equation*}
    so
    \begin{equation*} \cos(x + y) \left(1 + \frac{dy}{dx} \right) = 2 y \frac{dy}{dx} \cos x - y^2 \sin x. \end{equation*}
  2. Moving terms with factor \(\displaystyle\frac{dy}{dx}\) to the left, others to the right,
    \begin{equation*} \frac{dy}{dx}\cos(x+y) -2 y \frac{dy}{dx} \cos x = - y^2 \sin x - \cos(x+y) \end{equation*}
  3. Gathering the common factor \(\displaystyle\frac{dy}{dx}\) at left (and a factor \(-1\) at right),
    \begin{equation*} \frac{dy}{dx}[\cos(x+y) - 2 y \cos x] = - (y^2 \sin x + \cos(x + y)). \end{equation*}
  4. Dividing out,
    \begin{equation*} \frac{dy}{dx} = -\frac{y^2 \sin x + \cos(x + y)}{\cos(x + y) - 2 y \cos x}. \end{equation*}
Higher Derivatives: No Further Implicit Differentiation Needed!

Once a derivative has been found by implicit differentiation, you can compute the second and higher derivatives with no further implicit differentiation:

Solution

First, implicit differentiation gives \(\displaystyle \frac{dy}{dx} = -\frac{x^3}{y^3}\text{.}\)

Then the second derivative is \(\displaystyle \frac{d^2y}{dx^2} = \frac{d}{dx}\frac{dy}{dx} = -\frac{d}{dx}\frac{x^3}{y^3}\text{,}\) and the quotient rule combined with the hidden chain rule result

\begin{equation*} \frac{d(y^3)}{dx}=\frac{d(y^3)}{dy}\frac{dy}{dx} \end{equation*}

gives

\begin{equation*} \frac{d^2y}{dx^2} = - \frac{3x^2 y^3 - x^3 3y^2 dy/dx}{(y^3)^2}. \end{equation*}

Finally, inserting the above result for \(dy/dx\) gives

\begin{equation*} \frac{d^2y}{dx^2} = -\frac{3x^2 y^3 - x^3 3y^2 (-x^3/y^3)}{y^6}, = -\frac{3x^2 y^4 + 3 x^6 }{y^7}\text{.} \end{equation*}
Derivatives of the Inverse Trigonometric Functions.

We can compute the derivative of an inverse trigonometric function like \(y=\arcsin x = \sin^{-1}x\) by writing an equation involving the original “non-inverse” function whose derivative we know, like

\begin{equation*} x = \sin y\text{,} \end{equation*}

and then using implicit differentiation.

One thing we must be careful about is that no trigonometric function is invertible on its entire natural domain, so we limit the domain to make it satisfy the Horizontal Line Test. For example, the domain of \(\sin\) is restricted to \([-\pi/2,\pi/2]\text{,}\) so that \(\sin^{-1} x\) has range \([-\pi/2,\pi/2]\text{,}\) and domain \([-1,1]\text{.}\)

Implicit differentiation of \(x = \sin y\) gives \(\displaystyle 1 = \cos y \cdot \frac{dy}{dx}\text{,}\) so

\begin{equation*} \frac{dy}{dx}= \frac{1}{\cos y}, \quad \cos y \neq 0. \end{equation*}

But we want this as a function of \(x\text{,}\) not \(y\text{!}\)

Using the identity \(\sin^2 y + \cos^2 y = 1\) along with \(x=\sin y\text{,}\) we get \(\cos^2 y + x^2 = 1\text{,}\) so \(\cos y = \pm \sqrt{1-x^2}.\) The range of \(y\) values \([-\pi/2,\pi/2]\) ensures that \(\cos y\) is not negative, so \(\cos y = \sqrt{1-x^2}\) and

\begin{equation*} \frac{d}{dx} \arcsin x = \frac{1}{\sqrt{1-x^2}}, \quad -1 < x < 1. \end{equation*}

The restriction on the \(x\) values is to avoid division by zero, and shows that \(\arcsin x\) is not differentiable at the endpoints of its domain, due to “vertical tangents” (as with \(\sqrt{x}\) at \((0,0)\)). This happens because sine has horizontal tangents at the endpoints of its domain.

This happens quite often with inverse functions, because restricting the domain of a function to make it satisfy the Horizontal Line Test often requires ending the domain at a point where the tangent is horizontal. For example, the same happens when we cut off the domain of \(y=x^2\) at \(x=0\) to get its inverse \(\sqrt{x}\text{.}\) It also happens with the inverses of \(\cos\text{,}\) \(\sec\) and \(\csc\text{.}\)

The Derivatives of Other Inverse Trig. Functions.

Similar calculations give

\begin{gather*} \frac{d}{dx} \sin^{-1} x = \frac{1}{\sqrt{1-x^2}} \qquad \frac{d}{dx} \sec^{-1} x = \frac{1}{x\sqrt{x^2-1}}\\ \frac{d}{dx} \cos^{-1} x = -\frac{1}{\sqrt{1-x^2}} \qquad \frac{d}{dx} \csc^{-1} x = -\frac{1}{x\sqrt{x^2-1}}\\ \frac{d}{dx} \tan^{-1} x = \frac{1}{1+x^2} \qquad \frac{d}{dx} \cot^{-1} x = - \frac{1}{1+x^2} \end{gather*}
Differentiate \(\displaystyle y=\frac{1}{\sin^{-1}x}\text{,}\) being careful with the two uses of superscript -1!
Differentiate \(\displaystyle f(x)=x \arctan \sqrt{x}\text{,}\) being careful to distinguish products from compositions.

Recommended Exercises Recommended Exercises

Study Exercises 3, 5, 10, 15, 25, 29, 34, 35 and 51 from the text.